Optimal. Leaf size=154 \[ \frac {\tan (e+f x)}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 \tan (e+f x)}{15 a c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {8 \tan (e+f x)}{15 a^2 c^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]
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Rubi [A]
time = 0.11, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {3604, 40, 39}
\begin {gather*} \frac {8 \tan (e+f x)}{15 a^2 c^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}+\frac {4 \tan (e+f x)}{15 a c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {\tan (e+f x)}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 39
Rule 40
Rule 3604
Rubi steps
\begin {align*} \int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {1}{(a+i a x)^{7/2} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\tan (e+f x)}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 \text {Subst}\left (\int \frac {1}{(a+i a x)^{5/2} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=\frac {\tan (e+f x)}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 \tan (e+f x)}{15 a c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {8 \text {Subst}\left (\int \frac {1}{(a+i a x)^{3/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{15 a c f}\\ &=\frac {\tan (e+f x)}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 \tan (e+f x)}{15 a c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {8 \tan (e+f x)}{15 a^2 c^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}
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Mathematica [A]
time = 3.70, size = 117, normalized size = 0.76 \begin {gather*} -\frac {\sec ^2(e+f x) (\cos (3 (e+f x))+i \sin (3 (e+f x))) (150 \sin (e+f x)+25 \sin (3 (e+f x))+3 \sin (5 (e+f x))) \sqrt {c-i c \tan (e+f x)}}{240 a^2 c^3 f (-i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.35, size = 105, normalized size = 0.68
method | result | size |
derivativedivides | \(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (1+\tan ^{2}\left (f x +e \right )\right ) \tan \left (f x +e \right ) \left (8 \left (\tan ^{4}\left (f x +e \right )\right )+20 \left (\tan ^{2}\left (f x +e \right )\right )+15\right )}{15 f \,a^{3} c^{3} \left (\tan \left (f x +e \right )+i\right )^{4} \left (-\tan \left (f x +e \right )+i\right )^{4}}\) | \(105\) |
default | \(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (1+\tan ^{2}\left (f x +e \right )\right ) \tan \left (f x +e \right ) \left (8 \left (\tan ^{4}\left (f x +e \right )\right )+20 \left (\tan ^{2}\left (f x +e \right )\right )+15\right )}{15 f \,a^{3} c^{3} \left (\tan \left (f x +e \right )+i\right )^{4} \left (-\tan \left (f x +e \right )+i\right )^{4}}\) | \(105\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.63, size = 76, normalized size = 0.49 \begin {gather*} \frac {3 \, \sin \left (5 \, f x + 5 \, e\right ) + 25 \, \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (5 \, f x + 5 \, e\right ), \cos \left (5 \, f x + 5 \, e\right )\right )\right ) + 150 \, \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (5 \, f x + 5 \, e\right ), \cos \left (5 \, f x + 5 \, e\right )\right )\right )}{240 \, a^{\frac {5}{2}} c^{\frac {5}{2}} f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 1.08, size = 119, normalized size = 0.77 \begin {gather*} \frac {\sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-3 i \, e^{\left (12 i \, f x + 12 i \, e\right )} - 28 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 175 i \, e^{\left (8 i \, f x + 8 i \, e\right )} + 175 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 28 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i\right )} e^{\left (-5 i \, f x - 5 i \, e\right )}}{480 \, a^{3} c^{3} f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}} \left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 6.13, size = 163, normalized size = 1.06 \begin {gather*} \frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,125{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,22{}\mathrm {i}+\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}+175\,\sin \left (2\,e+2\,f\,x\right )+28\,\sin \left (4\,e+4\,f\,x\right )+3\,\sin \left (6\,e+6\,f\,x\right )-150{}\mathrm {i}\right )}{480\,a^3\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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